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Chapter 11: Problem 219

If the eccentricity of a hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{b^{2}}=1\), which passes through \((k, 2)\), is \(\frac{\sqrt{13}}{3}\), then the value of \(k^{2}\) is (a) 18 (b) 8 (c) 1 (d) 2

Short Answer

Expert verified
The correct answer is (a) 18.

Step by step solution

01

Identify Hyperbola Characteristics

The given hyperbola has the equation \( \frac{x^{2}}{9} - \frac{y^{2}}{b^{2}} = 1 \). This equation indicates that \( a^{2} = 9 \) and \( b^{2} \) is unknown. The eccentricity \( e \) given is \( \frac{\sqrt{13}}{3} \). Recall that the eccentricity of a hyperbola is given by \( e = \frac{\sqrt{a^{2} + b^{2}}}{a} \).
02

Express Eccentricity Relation

Substitute \( a^{2} = 9 \) and \( e = \frac{\sqrt{13}}{3} \) into the eccentricity formula \( e = \frac{\sqrt{a^{2} + b^{2}}}{a} \). Solve for \( b^{2} \): \( \frac{\sqrt{13}}{3} = \frac{\sqrt{9 + b^{2}}}{3} \).
03

Solve for \( b^2 \)

Square both sides and solve for \( b^2 \): \( 13 = 9 + b^{2} \Rightarrow b^{2} = 4 \).
04

Use Point \((k,2)\) to Find \(k\)

Since the hyperbola passes through the point \((k, 2)\), substitute \(x = k\) and \(y = 2\) into the hyperbola equation. Substitute \( b^{2} = 4 \) from Step 3 into the equation: \( \frac{k^{2}}{9} - \frac{2^{2}}{4} = 1 \).
05

Solve for \(k^2\)

Simplify the equation \( \frac{k^{2}}{9} - 1 = 1 \). Therefore, \( \frac{k^{2}}{9} = 2 \). Multiply both sides by 9 to get \( k^{2} = 18 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. They are fundamental in understanding various shapes in geometry. There are four main types of conics: circles, ellipses, parabolas, and hyperbolas. Each type is defined based on the angle and position of the intersecting plane.
  • A circle is formed when the plane cuts perpendicular to the cone's axis.
  • An ellipse results when the plane cuts the cone at an angle, not reaching the base.
  • A parabola is produced when the plane is parallel to the edge of the cone.
  • A hyperbola occurs when the plane intersects both nappes of the cone.
These shapes are described by specific equations in coordinate geometry and have unique properties such as eccentricity, which defines their shape.
Hyperbola
A hyperbola is a type of conic section characterized by its distinctive open curves. It is defined as the locus of points where the difference of the distances to two fixed points, called foci, remains constant.A hyperbola has two separate branches that diverge away from each other, unlike an ellipse, which is a closed curve.The standard equation of a hyperbola differs based on its orientation:
  • Horizontal Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
  • Vertical Hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
In the context of the given problem, the hyperbola equation is \( \frac{x^{2}}{9} - \frac{y^{2}}{b^{2}} = 1 \), which is a horizontal hyperbola with \( a^2 = 9 \). This equation helps determine other elements like eccentricity and points on the curve.
Eccentricity
Eccentricity is a crucial concept in understanding conic sections, specifically measuring how much a conic deviates from being circular. It is denoted by the letter "e."
  • For circles, \( e = 0 \)
  • For ellipses, \( 0 < e < 1 \)
  • For parabolas, \( e = 1 \)
  • For hyperbolas, \( e > 1 \)
The eccentricity of a hyperbola is calculated as \( e = \frac{\sqrt{a^2 + b^2}}{a} \). In this problem, the given eccentricity is \( \frac{\sqrt{13}}{3} \), which helps us find unknowns like \( b^2 \) by rearranging and solving the formula.
Coordinate Geometry
Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. It allows for precise and systematic solutions to geometrical problems by using algebraic equations.In coordinate geometry, we utilize equations like those of the conic sections to locate points, lines, and curves in a plane.
  • The Cartesian plane is made up of a horizontal x-axis and vertical y-axis.
  • Points are represented as coordinates \((x, y)\).
  • Equations describe paths or loci of points, such as lines through linear equations and curves through quadratic expressions.
For instance, in the hyperbola exercise, the point \((k, 2)\) is used to substitute into the equation to find unknown values, demonstrating how coordinate geometry can solve for specific values on a graph.

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Most popular questions from this chapter

If \(P\) and \(Q\) are the points of intersection of the circles \(x^{2}+y^{2}+3 x+7 y+2 p-5=0\) and \(x^{2}+y^{2}+2 x+2 y-p^{2}=0\) then there is a circle passing through \(\mathrm{P}, \mathrm{Q}\) and \((1,1)\) for: (a) all except one value of \(p\) (b) all except two values of \(p\) (c) exactly one value of \(p\) (d) all values of \(p\)

Let \(P\) be a point on the parabola, \(y^{2}=12 x\) and \(N\) be the foot of the perpendicular drawn from \(P\) on the axis of the parabola. A line is now drawn through the mid-point \(M\) of \(P N\), parallel to its axis which meets the parabola at \(Q\). If the \(y\)-intercept of the line \(N Q\) is \(\frac{4}{3}\), then : (a) \(P N=4\) (b) \(M Q=\frac{1}{3}\) (c) \(M Q=\frac{1}{4}\) (d) \(P N=3\)

If the circles \(x^{2}+y^{2}+5 K x+2 y+K=0\) and \(2\left(x^{2}+y^{2}\right)+\) \(2 \mathrm{Kx}+3 \mathrm{y}-1=0,(\mathrm{~K} \in \mathbf{R})\), intersect at the points \(\mathrm{P}\) and \(\mathrm{Q}\), then the line \(4 x+5 y-K=0\) passes through \(P\) and \(Q\), for: (a) infinitely many values of \(\mathrm{K}\) (b) no value of \(\mathrm{K}\). (c) exactly two values of \(\mathrm{K}\) (d) exactly one value of \(\mathrm{K}\)

The equation \(\operatorname{Im}\left(\frac{i z-2}{z-i}\right)+1=0, z \in C, z \neq i\) represents a part of a circle having radius equal to : (a) 2 (b) 1 (c) \(\frac{3}{4}\) (d) \(\frac{1}{2}\)

Let \(P Q\) be a diameter of the circle \(x^{2}+y^{2}=9\). If \(\alpha\) and \(\beta\) are the lengths of the perpendiculars from \(P\) and \(Q\) on the straight line, \(x+y=2\) respectively, then the maximum value of \(\alpha \beta\) is ___ .
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